91Ó°ÊÓ

In calculus, continuity refers to a fundamental property of functions, dictating how they behave at various points and over different intervals. For a function to be continuous at a certain point, three conditions must be met: the function must be defined at the point, it must have a limit as it approaches the point, and the value of the function at that point must equal the limit.

When analyzing the function in the exercise, one can infer that it has to be analyzed at all points in its domain to establish intervals where it is continuous. The function is continuous on its domain, shown by smooth, unbroken graphs. When a function fails to meet any of the continuity conditions at a point, it's said to have a discontinuity there. Understanding where these breakpoints occur is crucial because it influences functions like integration and differentiability.

Rational functions are ratios of two polynomials, represented as \(f(x) = \frac{p(x)}{q(x)}\), where both \(p(x)\) and \(q(x)\) are polynomials and \(q(x) \eq 0\). The function given in the exercise, \(f(x)=\frac{1}{x^{2}-9}\), is a simple rational function.

Rational functions are continuous wherever their denominators are non-zero. This is because division by zero is undefined, leading to points or intervals of discontinuity. Distinct from polynomials, the domains of rational functions can exclude certain real numbers (values of \(x\) for which \(q(x) = 0\)). By identifying and excluding these values, we find the function's domain and the intervals of its continuity.

Identifying discontinuities in a function is key in understanding its behavior. Discontinuities may occur where a function does not exist (point discontinuity), where there is a sudden jump (jump discontinuity), or where the function heads off towards infinity (infinite discontinuity).

In the exercise, the discontinuities of the rational function \(f(x)\) occur when the denominator is zero, which is when \(x^2 - 9 = 0\). Solving this equation reveals the values \(x = -3\) and \(x = 3\), the points of infinite discontinuities due to division by zero. By excluding these points, we find that \(f(x)\) is continuous in the intervals \( (-\infty, -3), (-3, 3), \) and \( (3, \infty) \).

Quadratic equations are second-degree polynomial equations of the form \(ax^2 + bx + c = 0\). These equations can have two real solutions, one real solution, or no real solutions, depending on the discriminant \(b^2 - 4ac\).

The step by step solution for the exercise involved solving the quadratic equation \(x^2 - 9 = 0\) to find the values that cause the function to be undefined. Factoring yields \( (x + 3)(x - 3) = 0 \) leading to the solutions \(x = -3\) and \(x = 3\). Solving quadratic equations is crucial in finding potential discontinuities in rational functions, as these solutions indicate where the function cannot exist or operates unexpectedly.